Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int getHighestBit(int a) {
  for (int i = 31; i >= 0; i--) {
    if (a >> i & 1) {
      return 1 << i;
    }
  }
  return 0;
}
vector<int> spaceVector(vector<int> a, vector<pair<int, int> > &operation) {
  int n = ((int)(a).size());
  int start = 0;
  while (start < n) {
    int ind = start;
    int maxi = a[start];
    for (int j = start + 1; j < n; j++) {
      if (maxi < a[j]) {
        maxi = a[j];
        ind = j;
      }
    }
    if (ind != start) {
      swap(a[ind], a[start]);
      operation.push_back(make_pair(ind, start));
      operation.push_back(make_pair(start, ind));
      operation.push_back(make_pair(ind, start));
    }
    int higehstBit = getHighestBit(a[start]);
    if (higehstBit == 0) {
      break;
    }
    for (int j = start + 1; j < n; j++) {
      if ((a[j] & higehstBit) != 0) {
        a[j] ^= a[start];
        operation.push_back(make_pair(j, start));
      }
    }
    start++;
  }
  a.resize(start);
  return a;
}
int main() {
  vector<int> x, y;
  int n;
  cin >> n;
  x.resize(n);
  y.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> y[i];
  }
  vector<pair<int, int> > opGo, opRev;
  x = spaceVector(x, opGo);
  y = spaceVector(y, opRev);
  bool can = true;
  if (((int)(x).size()) < ((int)(y).size())) {
    can = false;
  } else {
    int yIter = 0;
    int xIter = 0;
    int step = 0;
    while (can && yIter < ((int)(y).size())) {
      int higehstBit = getHighestBit(y[yIter]);
      while (step < ((int)(x).size()) && getHighestBit(x[step]) > higehstBit) {
        step++;
      }
      if (step >= ((int)(x).size()) || getHighestBit(x[step]) != higehstBit) {
        can = false;
        break;
      }
      if (xIter != step) {
        swap(x[xIter], x[step]);
        opGo.push_back(make_pair(xIter, step));
        opGo.push_back(make_pair(step, xIter));
        opGo.push_back(make_pair(xIter, step));
      }
      step++;
      int nextSteps = step;
      while (x[xIter] != y[yIter]) {
        int dif = x[xIter] ^ y[yIter];
        higehstBit = getHighestBit(dif);
        while (nextSteps < ((int)(x).size()) &&
               getHighestBit(x[nextSteps]) > higehstBit) {
          nextSteps++;
        }
        if (nextSteps >= ((int)(x).size()) ||
            getHighestBit(x[nextSteps]) != higehstBit) {
          can = false;
          break;
        }
        x[xIter] ^= x[nextSteps];
        opGo.push_back(make_pair(xIter, nextSteps));
      }
      xIter++;
      yIter++;
    }
  }
  if (!can) {
    cout << "-1" << endl;
  } else {
    for (int i = ((int)(y).size()); i < ((int)(x).size()); i++) {
      opGo.push_back(make_pair(i, i));
    }
    cout << ((int)(opRev).size()) + ((int)(opGo).size()) << endl;
    reverse(opRev.begin(), opRev.end());
    for (int i = 0; i < ((int)(opGo).size()); i++) {
      cout << opGo[i].first + 1 << ' ' << opGo[i].second + 1 << endl;
    }
    for (int i = 0; i < ((int)(opRev).size()); i++) {
      cout << opRev[i].first + 1 << ' ' << opRev[i].second + 1 << endl;
    }
  }
  return 0;
}